If a unit vector $$\vec{a}$$ makes angles $$\frac{\pi}{3}$$ with $$\hat{i}$$, $$\frac{\pi}{4}$$ with $$\hat{j}$$ and $$\theta \in (0, \pi)$$ with $$\hat{k}$$, then a value of $$\theta$$ is:

For any vector in three-dimensional space, if the angles made with the positive coordinate axes are $$\alpha,\;\beta,\;\gamma,$$ then (direction-cosine identity)

$$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma = 1.$$

Here the given angles are $$\alpha=\dfrac{\pi}{3},\;\beta=\dfrac{\pi}{4},\;\gamma=\theta,$$ so we substitute them into the identity.

First we evaluate each cosine:

$$\cos\dfrac{\pi}{3} = \dfrac12 \quad\Longrightarrow\quad \cos^{2}\dfrac{\pi}{3}=\left(\dfrac12\right)^{2}=\dfrac14.$$

$$\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}\quad\Longrightarrow\quad \cos^{2}\dfrac{\pi}{4}=\left(\dfrac{\sqrt2}{2}\right)^{2}=\dfrac12.$$

Now we place these values in the identity and solve for $$\cos^{2}\theta$$:

$$\cos^{2}\dfrac{\pi}{3}+\cos^{2}\dfrac{\pi}{4}+\cos^{2}\theta = 1$$

$$\Longrightarrow\quad \dfrac14+\dfrac12+\cos^{2}\theta = 1.$$

We add the fractions on the left:

$$\dfrac14+\dfrac12 = \dfrac14+\dfrac24 = \dfrac34.$$

So we have

$$\dfrac34+\cos^{2}\theta = 1.$$

Subtracting $$\dfrac34$$ from both sides gives

$$\cos^{2}\theta = 1-\dfrac34 = \dfrac14.$$

Taking the square root, we obtain

$$\cos\theta = \pm\dfrac12.$$

Because $$\theta\in(0,\pi),$$ both values are possible in principle, yielding

$$\theta = \dfrac{\pi}{3}\quad\text{or}\quad\theta = \dfrac{2\pi}{3}.$$

Among the options provided, only $$\dfrac{2\pi}{3}$$ appears.

Hence, the correct answer is Option D.